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3 years ago

Parallel to y=-1/4x-12 through (12,10)

Mathematics

1 answer:

Yanka [14]3 years ago

8 0

Y=-1/4x+13 not sure though

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Evaluate the expression for s=4 and t=8

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I am confused please can you help me

lorasvet [3.4K]

Domain x|x = -2, 2
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3 years ago

X ( y - x y ), where x= -5 and y= -2

adell [148]

Answer:

Step-by-step explanation:

-5(-2-10)

according to pemdas...solve parenthesis first

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= 60

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3 years ago

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Solve the following systems of equations using the matrix method: a. 3x1 + 2x2 + 4x3 = 5 2x1 + 5x2 + 3x3 = 17 7x1 + 2x2 + 2x3 =

lara [203]

Answer:

a. The solutions are

$\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\begin{pmatrix}\frac{11}{13}\\ \frac{50}{13}\\ -\frac{17}{13}\end{pmatrix}$

b. The solutions are

$\left[\begin{array}{c}x&y&z\\\end{array}\right]=\begin{pmatrix}\frac{54}{235}\\ \frac{6}{47}\\ \frac{24}{235}\end{pmatrix}$

c. The solutions are

$\left[\begin{array}{c}x_1&x_2&x_3&x_4\\\end{array}\right]=\begin{pmatrix}\frac{22}{9}\\ \frac{164}{9}\\ \frac{139}{9}\\ -\frac{37}{3}\end{pmatrix}$

Step-by-step explanation:

Solving a system of linear equations using matrix method, we may define a system of equations with the same number of equations as variables as:

$A\cdot X=B$

where X is the matrix representing the variables of the system, B is the matrix representing the constants, and A is the coefficient matrix.

Then the solution is this:

$X=A^{-1}B$

a. Given the system:

$3x_1 + 2x_2 + 4x_3 = 5 \\2x_1 + 5x_2 + 3x_3 = 17 \\7x_1 + 2x_2 + 2x_3 = 11$

The coefficient matrix is:

$A=\left[\begin{array}{ccc}3&2&4\\2&5&3\\7&2&2\end{array}\right]$

The variable matrix is:

$X=\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]$

The constant matrix is:

$B=\left[\begin{array}{c}5&17&11\\\end{array}\right]$

First, we need to find the inverse of the A matrix. To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

$\left[ \begin{array}{ccc|ccc}3&2&4&1&0&0 \\\\ 2&5&3&0&1&0 \\\\ 7&2&2&0&0&1\end{array}\right]$

This matrix can be transformed by a sequence of elementary row operations to the matrix

$\left[ \begin{array}{ccc|ccc}1&0&0&- \frac{2}{39}&- \frac{2}{39}&\frac{7}{39} \\\\ 0&1&0&- \frac{17}{78}&\frac{11}{39}&\frac{1}{78} \\\\ 0&0&1&\frac{31}{78}&- \frac{4}{39}&- \frac{11}{78}\end{array}\right]$

And the inverse of the A matrix is

$A^{-1}=\left[ \begin{array}{ccc} - \frac{2}{39} & - \frac{2}{39} & \frac{7}{39} \\\\ - \frac{17}{78} & \frac{11}{39} & \frac{1}{78} \\\\ \frac{31}{78} & - \frac{4}{39} & - \frac{11}{78} \end{array} \right]$

Next, multiply $A^ {-1}$ by $B$

$X=A^{-1}\cdot B$

$\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\left[ \begin{array}{ccc} - \frac{2}{39} & - \frac{2}{39} & \frac{7}{39} \\\\ - \frac{17}{78} & \frac{11}{39} & \frac{1}{78} \\\\ \frac{31}{78} & - \frac{4}{39} & - \frac{11}{78} \end{array} \right] \cdot \left[\begin{array}{c}5&17&11\end{array}\right]$

$\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\begin{pmatrix}-\frac{2}{39}&-\frac{2}{39}&\frac{7}{39}\\ -\frac{17}{78}&\frac{11}{39}&\frac{1}{78}\\ \frac{31}{78}&-\frac{4}{39}&-\frac{11}{78}\end{pmatrix}\begin{pmatrix}5\\ 17\\ 11\end{pmatrix}=\begin{pmatrix}\frac{11}{13}\\ \frac{50}{13}\\ -\frac{17}{13}\end{pmatrix}$

The solutions are

$\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\begin{pmatrix}\frac{11}{13}\\ \frac{50}{13}\\ -\frac{17}{13}\end{pmatrix}$

b. To solve this system of equations

$x -y - z = 0 \\30x + 40y = 12 \\30x + 50z = 12$

The coefficient matrix is:

$A=\left[\begin{array}{ccc}1&-1&-1\\30&40&0\\30&0&50\end{array}\right]$

The variable matrix is:

$X=\left[\begin{array}{c}x&y&z\\\end{array}\right]$

The constant matrix is:

$B=\left[\begin{array}{c}0&12&12\\\end{array}\right]$

The inverse of the A matrix is

$A^{-1}=\left[ \begin{array}{ccc} \frac{20}{47} & \frac{1}{94} & \frac{2}{235} \\\\ - \frac{15}{47} & \frac{4}{235} & - \frac{3}{470} \\\\ - \frac{12}{47} & - \frac{3}{470} & \frac{7}{470} \end{array} \right]$

The solutions are

$\left[\begin{array}{c}x&y&z\\\end{array}\right]=\begin{pmatrix}\frac{54}{235}\\ \frac{6}{47}\\ \frac{24}{235}\end{pmatrix}$

c. To solve this system of equations

$4x_1 + 2x_2 + x_3 + 5x_4 = 0 \\3x_1 + x_2 + 4x_3 + 7x_4 = 1\\ 2x_1 + 3x_2 + x_3 + 6x_4 = 1 \\3x_1 + x_2 + x_3 + 3x_4 = 4\\$

The coefficient matrix is:

$A=\left[\begin{array}{cccc}4&2&1&5\\3&1&4&7\\2&3&1&6\\3&1&1&3\end{array}\right]$

The variable matrix is:

$X=\left[\begin{array}{c}x_1&x_2&x_3&x_4\\\end{array}\right]$

The constant matrix is:

$B=\left[\begin{array}{c}0&1&1&4\\\end{array}\right]$

The inverse of the A matrix is

$A^{-1}=\left[ \begin{array}{cccc} - \frac{1}{9} & - \frac{1}{9} & - \frac{1}{9} & \frac{2}{3} \\\\ - \frac{32}{9} & - \frac{5}{9} & \frac{13}{9} & \frac{13}{3} \\\\ - \frac{28}{9} & - \frac{1}{9} & \frac{8}{9} & \frac{11}{3} \\\\ \frac{7}{3} & \frac{1}{3} & - \frac{2}{3} & -3 \end{array} \right]$

The solutions are

$\left[\begin{array}{c}x_1&x_2&x_3&x_4\\\end{array}\right]=\begin{pmatrix}\frac{22}{9}\\ \frac{164}{9}\\ \frac{139}{9}\\ -\frac{37}{3}\end{pmatrix}$

7 0

3 years ago

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Hunter owns a food truck that sells tacos and burritos. He sells each taco for $3.25

WITCHER [35]

Answer:

y = 40 x = 80

Step-by-step explanation:

520 = 3.25x + 6.5y x+y=120

x+y=120

x = 120-y

520 = 3.25(120-y)+6.5y

520 = 390-3.25y+6.5y

520 = 390+3.25y

130=3.25y

Divide each side by 3.25

40 = y

x = 120-y

x = 120-(40)

x = 80

3 0

3 years ago